Question

If sum of first n terms of an A.P. is (3n$^2$ – 2n) ∀ n ∈ N, then the value of $\sum_{n=1}^{\infty} \frac{21}{(S_nS_{n+2}+S_{n-1}S_{n+1}) - (S_nS_{n+1}+S_{n-1}S_{n+2})}$ is

Answer 1

2

Answer 2

The sum of the first n terms of an A.P. is given by $S_n = 3n^2 - 2n$ for all $n \in N$. The n-th term of the A.P. is $a_n = S_n - S_{n-1}$ for $n > 1$, and $a_1 = S_1$.

$S_1 = 3(1)^2 - 2(1) = 1$. So $a_1 = 1$.

For $n > 1$, $S_{n-1} = 3(n-1)^2 - 2(n-1) = 3(n^2 - 2n + 1) - 2n + 2 = 3n^2 - 6n + 3 - 2n + 2 = 3n^2 - 8n + 5$.

$a_n = (3n^2 - 2n) - (3n^2 - 8n + 5) = 6n - 5$.

This formula also works for $n=1$: $a_1 = 6(1) - 5 = 1$. So the general term of the A.P. is $a_n = 6n - 5$. The common difference of the A.P. is $d = a_{n+1} - a_n = (6(n+1) - 5) - (6n - 5) = 6n + 6 - 5 - 6n + 5 = 6$.

The expression in the denominator of the sum term is $(S_nS_{n+2}+S_{n-1}S_{n+1}) - (S_nS_{n+1}+S_{n-1}S_{n+2})$. Let's simplify this expression: $D_n = S_nS_{n+2}+S_{n-1}S_{n+1} - S_nS_{n+1}-S_{n-1}S_{n+2}$ $D_n = (S_nS_{n+2} - S_nS_{n+1}) - (S_{n-1}S_{n+2} - S_{n-1}S_{n+1})$ $D_n = S_n(S_{n+2} - S_{n+1}) - S_{n-1}(S_{n+2} - S_{n+1})$ $D_n = (S_n - S_{n-1})(S_{n+2} - S_{n+1})$.

We know that $S_k - S_{k-1} = a_k$ for $k \ge 2$. So $S_n - S_{n-1} = a_n$ for $n \ge 2$. And $S_{n+2} - S_{n+1} = a_{n+2}$ for $n+2 \ge 2$, which means $n \ge 0$. Since $n \in N$, this holds for $n \ge 1$. For $n \ge 2$, $D_n = a_n a_{n+2}$.

For $n=1$, the expression involves $S_{n-1} = S_0$. The formula $S_n = 3n^2 - 2n$ gives $S_0 = 3(0)^2 - 2(0) = 0$. Assuming $S_0=0$, the denominator for $n=1$ is: $D_1 = (S_1S_3+S_0S_2) - (S_1S_2+S_0S_3) = (S_1S_3+0) - (S_1S_2+0) = S_1(S_3 - S_2)$. $S_1 = 1$. $a_2 = 6(2) - 5 = 7$. $a_3 = 6(3) - 5 = 13$. $S_2 = S_1 + a_2 = 1 + 7 = 8$. $S_3 = S_2 + a_3 = 8 + 13 = 21$. $D_1 = S_1(S_3 - S_2) = 1(21 - 8) = 1 \times 13 = 13$. The formula $a_n a_{n+2}$ for $n=1$ gives $a_1 a_3 = 1 \times 13 = 13$. Thus, the denominator is $D_n = a_n a_{n+2}$ for all $n \ge 1$.

The sum is $\sum_{n=1}^{\infty} \frac{21}{a_n a_{n+2}}$. Substitute $a_n = 6n - 5$ and $a_{n+2} = 6(n+2) - 5 = 6n + 12 - 5 = 6n + 7$. The term is $\frac{21}{(6n-5)(6n+7)}$.

We use partial fraction decomposition: $\frac{21}{(6n-5)(6n+7)} = \frac{A}{6n-5} + \frac{B}{6n+7}$. $21 = A(6n+7) + B(6n-5)$. Setting $6n-5=0 \implies n=5/6$, we get $21 = A(6(5/6)+7) = A(5+7) = 12A \implies A = 21/12 = 7/4$. Setting $6n+7=0 \implies n=-7/6$, we get $21 = B(6(-7/6)-5) = B(-7-5) = -12B \implies B = 21/(-12) = -7/4$. So $\frac{21}{(6n-5)(6n+7)} = \frac{7/4}{6n-5} - \frac{7/4}{6n+7} = \frac{7}{4} \left( \frac{1}{6n-5} - \frac{1}{6n+7} \right)$. Recognizing $a_n = 6n-5$ and $a_{n+2} = 6n+7$, the term is $\frac{7}{4} \left( \frac{1}{a_n} - \frac{1}{a_{n+2}} \right)$.

The sum is $\sum_{n=1}^{\infty} \frac{7}{4} \left( \frac{1}{a_n} - \frac{1}{a_{n+2}} \right)$. This is a telescoping series. Let's consider the partial sum $S_N = \sum_{n=1}^{N} \frac{7}{4} \left( \frac{1}{a_n} - \frac{1}{a_{n+2}} \right)$. $S_N = \frac{7}{4} \left[ \left(\frac{1}{a_1} - \frac{1}{a_3}\right) + \left(\frac{1}{a_2} - \frac{1}{a_4}\right) + \left(\frac{1}{a_3} - \frac{1}{a_5}\right) + \dots + \left(\frac{1}{a_{N-1}} - \frac{1}{a_{N+1}}\right) + \left(\frac{1}{a_N} - \frac{1}{a_{N+2}}\right) \right]$. The terms cancel out, leaving: $S_N = \frac{7}{4} \left[ \frac{1}{a_1} + \frac{1}{a_2} - \frac{1}{a_{N+1}} - \frac{1}{a_{N+2}} \right]$.

To find the sum of the infinite series, we take the limit as $N \to \infty$: $\sum_{n=1}^{\infty} \frac{21}{a_n a_{n+2}} = \lim_{N \to \infty} S_N = \frac{7}{4} \lim_{N \to \infty} \left[ \frac{1}{a_1} + \frac{1}{a_2} - \frac{1}{a_{N+1}} - \frac{1}{a_{N+2}} \right]$. As $N \to \infty$, $a_{N+1} = 6(N+1) - 5 = 6N+1 \to \infty$ and $a_{N+2} = 6(N+2) - 5 = 6N+7 \to \infty$. So $\lim_{N \to \infty} \frac{1}{a_{N+1}} = 0$ and $\lim_{N \to \infty} \frac{1}{a_{N+2}} = 0$. The sum is $\frac{7}{4} \left( \frac{1}{a_1} + \frac{1}{a_2} - 0 - 0 \right) = \frac{7}{4} \left( \frac{1}{1} + \frac{1}{7} \right)$.

We have $a_1 = 1$ and $a_2 = 7$. The sum is $\frac{7}{4} \left( \frac{1}{1} + \frac{1}{7} \right) = \frac{7}{4} \left( 1 + \frac{1}{7} \right) = \frac{7}{4} \left( \frac{7+1}{7} \right) = \frac{7}{4} \times \frac{8}{7} = \frac{8}{4} = 2$.

The final answer is 2.

Explanation of the solution:

1. Find the general term $a_n$ of the A.P. from the given sum formula $S_n$. $a_n = S_n - S_{n-1}$ for $n>1$ and $a_1=S_1$.
2. Simplify the denominator of the term in the sum. It simplifies to $(S_n - S_{n-1})(S_{n+2} - S_{n+1})$.
3. Use the relation $S_k - S_{k-1} = a_k$ to express the denominator in terms of the terms of the A.P. The denominator is $a_n a_{n+2}$ for $n \ge 1$ (assuming $S_0=0$, which is consistent with the formula for $S_n$).
4. Substitute the formula for $a_n$ into the term $\frac{21}{a_n a_{n+2}}$.
5. Use partial fraction decomposition to rewrite the term $\frac{21}{a_n a_{n+2}}$ in the form $C (\frac{1}{a_n} - \frac{1}{a_{n+2}})$.
6. Evaluate the infinite sum using the telescoping series property. The partial sum $\sum_{n=1}^N (\frac{1}{a_n} - \frac{1}{a_{n+2}})$ simplifies to $\frac{1}{a_1} + \frac{1}{a_2} - \frac{1}{a_{N+1}} - \frac{1}{a_{N+2}}$.
7. Take the limit of the partial sum as $N \to \infty$. Since $a_n \to \infty$ as $n \to \infty$, the terms $\frac{1}{a_{N+1}}$ and $\frac{1}{a_{N+2}}$ go to 0.
8. Substitute the values of $a_1$ and $a_2$ to get the final sum.